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            <h1 id="seo-header">『算法-ACM竞赛-动态规划』DP60题-2HDU1025ConstructingRoadsInJGShining&#39;sKingdo</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-动态规划』DP60-题-2HDU1025ConstructingRoadsInJGShining’sKingdo"><a href="#『算法-ACM-竞赛-动态规划』DP60-题-2HDU1025ConstructingRoadsInJGShining’sKingdo" class="headerlink" title="『算法-ACM 竞赛-动态规划』DP60 题-2HDU1025ConstructingRoadsInJGShining’sKingdo"></a>『算法-ACM 竞赛-动态规划』DP60 题-2HDU1025ConstructingRoadsInJGShining’sKingdo</h1><h1 id="DP-60-题-2-HDU1025-Constructing-Roads-In-JGShining’s-Kingdom"><a href="#DP-60-题-2-HDU1025-Constructing-Roads-In-JGShining’s-Kingdom" class="headerlink" title="DP 60 题 -2 HDU1025 Constructing Roads In JGShining’s Kingdom"></a>DP 60 题 -2 HDU1025 Constructing Roads In JGShining’s Kingdom</h1><p>Problem Description</p>
<p>JGShining’s kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines.</p>
<p>Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource.</p>
<p>With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they’re unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don’t wanna build a road with other poor ones, and rich ones also can’t abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.</p>
<p>Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.</p>
<p>The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities … And so as the poor ones.</p>
<p>But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.</p>
<p>For example, the roads in Figure I are forbidden.</p>
<p><img src="https://imgconvert.csdnimg.cn/aHR0cDovL2FjbS5oZHUuZWR1LmNuL2RhdGEvaW1hZ2VzLzEwMjUtMS5wbmc?x-oss-process=image/format,png" srcset="/img/loading.gif" lazyload></p>
<p>In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^</p>
<p>Input</p>
<p>Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.</p>
<p>Output</p>
<p>For each test case, output the result in the form of sample.<br>You should tell JGShining what’s the maximal number of road(s) can be built.</p>
<p>Sample Input</p>
<pre><code class="hljs"> 
</code></pre>
<p>2 1 2 2 1 3 1 2 2 3 3 1</p>
<p>Sample Output</p>
<pre><code class="hljs"> 
</code></pre>
<p>Case 1: My king, at most 1 road can be built. Case 2: My king, at most 2 roads can be built.</p>
<p><em>Hint</em></p>
<p>Huge input, scanf is recommended._ _</p>
<p>Author</p>
<p>JGShining（极光炫影）</p>
<p>Recommend</p>
<p>We have carefully selected several similar problems for you: <a target="_blank" rel="noopener" href="http://acm.hdu.edu.cn/showproblem.php?pid=1024">1024</a> <a target="_blank" rel="noopener" href="http://acm.hdu.edu.cn/showproblem.php?pid=1081">1081</a> <a target="_blank" rel="noopener" href="http://acm.hdu.edu.cn/showproblem.php?pid=1074">1074</a> <a target="_blank" rel="noopener" href="http://acm.hdu.edu.cn/showproblem.php?pid=1080">1080</a> <a target="_blank" rel="noopener" href="http://acm.hdu.edu.cn/showproblem.php?pid=1114">1114</a></p>
<p>这题傻逼，我日了，题意很简单，就是不能交叉的连线最多有多少条，按照第一个排序后，直接 LIS，然后我觉得可能超时了，我觉得排序都可以省了，这里有个骚操作，因为他的 POOR 是连续的，所以可以</p>
<pre><code class="hljs">scanf(&quot;%d%d&quot;, &amp;a, &amp;b);
            a[a] = b;
</code></pre>
<p>省的排序了，然后想了想，LIS 还有 O（NlogN）复杂度，所以换了，然后 WA 了。我日我交了三遍都错了，什么原理呢，我觉得我的思路一点没错，然后看了博客，麻蛋，卡单复数，还卡回车，奶奶的！这谁顶得住！</p>
<p><img src="https://img-blog.csdnimg.cn/20190919004914465.png" srcset="/img/loading.gif" lazyload></p>
<pre><code class="hljs">#define _CRT_SECURE_NO_WARNINGS
#include&lt;iostream&gt;
#include&lt;queue&gt;
#include&lt;algorithm&gt;
#include&lt;set&gt;
#include&lt;cmath&gt;
#include&lt;vector&gt;
#include&lt;map&gt;
#include&lt;stack&gt;
#include&lt;bitset&gt;
#include&lt;cstdio&gt;
#include&lt;cstring&gt;
#define Swap(a,b) a^=b^=a^=b
#define cini(n) scanf(&quot;%d&quot;,&amp;n)
#define cinl(n) scanf(&quot;%lld&quot;,&amp;n)
#define cinc(n) scanf(&quot;%c&quot;,&amp;n)
#define cins(s) scanf(&quot;%s&quot;,s)
#define coui(n) printf(&quot;%d&quot;,n)
#define couc(n) printf(&quot;%c&quot;,n)
#define coul(n) printf(&quot;%lld&quot;,n)
#define speed ios_base::sync_with_stdio(0)
#define Max(a,b) a&gt;b?a:b
#define Min(a,b) a&lt;b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define INF  0x3f3f3f3f
#define maxn  100010
#define esp  1e-9
#define mp(a,b) make_pair(a,b)
using namespace std;
typedef long long ll;
//-----------------------*******----------------------------//
int a[1000000];
int dp[1000000];
int main() &#123;
    int n;
    int cse = 1;
    while(scanf(&quot;%d&quot;, &amp;n) != EOF) &#123;
        int i;
        int a, b;
        int len, pos;
        int val;
        memset(a, 0, sizeof(a));
        for(i = 0; i &lt; n; i++) &#123;
            scanf(&quot;%d%d&quot;, &amp;a, &amp;b);
            a[a] = b;
        &#125;
        len = 1;
        dp[0] =INF;
        for(i = 1; i &lt;=n; i++) &#123;
            val = a[i];
            if(val &gt; dp[len-1]) &#123;
                dp[len++] = val;
            &#125;
            else &#123;
                pos = upper_bound(dp,dp+len,val)-dp;
                dp[pos] = val;
            &#125;
        &#125;
        printf(&quot;Case %d:\n&quot;, cse++);
        printf(&quot;My king, at most %d %s can be built.\n\n&quot;, len, len&gt;1?&quot;roads&quot;:&quot;road&quot;);
    &#125;
    return 0;
&#125;
</code></pre>

                
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